2u Mathematics Marathon v1.0 (1 Viewer)

[KeypadSDM]

A and B are two points 200 metres apart. For what values of L is it possible to find a circular arc AB of length L metres? Justify your answer.

Spoiler

200 m < L

Say the location of the points A & B in the cartesian plane are:
A = (-100,0)
B = (100,0)

Then place the centre of the circle which produces the arc between A & B on the line x = 0, with coordinate: C = (0,k)

Note that the arc is drawn counterclockwise from B to A.

As k -> -oo, L -> 200 m
As k -> oo, L -> oo

And for any given length, you can find a location of C which will give that arc length.

However, if we only consider the minor arc [I.e. if C is above the x-axis, then we draw it clockwise from B to A], then we can state:

200 m < L <= 100pi m

I don't know if that's right, it seems too simple.

Next Question: [Another InTeGrAtIoN!]

Integral[from -pi/3 to pi/3] (sin[x] - 2x)/(1 + e^x2)

Trust me, it's a 2 unit integration, no doubt.

 

[icycloud]

Lol yeh it's right, it's meant to be simple: remember it's 2U!
Anyway the integral:

Spoiler

To cut a long story short, F(-x) = -F(x), thus F(x) is odd.
Due to symmetry, I = 0

Next Q:
Prove Sin[2x] = (Cos[pi/4 - x] + Sin[pi/4 - x])(Cos[pi/4 - x] - Sin[pi/4 - x])

 

[Templar]

Spoiler

(Cos[pi/4 - x] + Sin[pi/4 - x])(Cos[pi/4 - x] - Sin[pi/4 - x])
=cos^2[pi/4-x]-sin^2[pi/4-x]
=1-2sin^2[pi/4-x]
=1-2(sin[pi/4]cos[x]-cos[pi/4]sin[x])^2
=1-2(1/sqrt[2] (cos[x]-sin[x]))^2
=1-(cos^2[x]+sin^2[x]-2sin[x]cos[x])
=sin[2x]

 

[KeypadSDM]

Slightly easier way

Spoiler

Sin[2x]
=Cos[pi/2 - 2x]
=Cos²[pi/4 - x] - Sin²[pi/4 - x]
=(Cos[pi/4 - x] - Sin[pi/4 - x])*(Cos[pi/4 - x] + Sin[pi/4 - x])
As required

Seriously Templar...

 

[Jono_2007]

Post the next question up then!

 

[icycloud]

Post the next question up then!

Lol Jono, just for you:

Next question:
Two particles A and B start moving on the x axis at time t = 0. The position of particle A at time t is given by:

x = -6 + 2t - (1/2)t²

and the position of particle B at time t is given by:

x = 4sin(t).

Show that there are exactly two occassions, t₁ and t₂, when these particles have the same velocity. Furthermore, show that the distance travelled by particle A between these two occassions is:

4 - 2(t₁+t₂) + (1/2)(t₁²+t₂²)

Then, show that the two particles never meet.

 

[Jono_2007]

How the hell do you find that! Oh wait do you use integration! i think i know what to do. i might have a go!

 

[Jono_2007]

Nah, i cant work it out. So my above post is invalid. ok someone else try it.

 

[insert-username]

This is Applications of Calculus to the Physical World. I *think* the velocity of the particles is given by the derivative of their positions, so you derive the equations and equate them together. You should get two results - two times. Subtract the lower from the higher to get the time between when the particles have the same velocity, and substitute that value into the position of particle A to get how far it's travelled. Now, to go off and actually learn how to do that...


I_F

 

[Jono_2007]

This is the 1995 Maths 2/3 unit common paper, question 10. If only i had those solutions.

 

[Mountain.Dew]

Lol Jono, just for you:

Next question:
Two particles A and B start moving on the x axis at time t = 0. The position of particle A at time t is given by:

x = -6 + 2t - (1/2)t²

and the position of particle B at time t is given by:

x = 4sin(t).

Show that there are exactly two occassions, t₁ and t₂, when these particles have the same velocity. Furthermore, show that the distance travelled by particle A between these two occassions is:

4 - 2(t₁+t₂) + (1/2)(t₁²+t₂²)

Then, show that the two particles never meet.

im not going to do the question, (since it is reserved for jono_2007) but this is what i would do:

first, derive the velocity equation of both particles. graph their velocities on a velocity-time graph, and observe that there are two intersecting pts --> the two occasions, t₂ and t₁.

now, i observed that the curve x = -6 + 2t - (1/2)t² is negative definite, since a < 0 AND discriminant < 0. now, to find the max. TP of that curve...which happens to be t=2.

SO, the distance particle A travels between t₂ and t₁ IS the distance travelled between t₁ and t=2, AND distance travelled between t=2 AND t₂. the distance u should get is
4 - 2(t₁+t₂) + (1/2)(t₁²+t₂²).

the last part, where u prove that they dont meet --> i would imagine that u would equate t1=t2, find where they CAN meet, sub that value for t into x=4sint, and show that the two distances xB and xA are not equal. hence, they cant meet. not sure on this one, should be another way of doing it.

 

[KeypadSDM]

Bah, reserved = Boring.

Long question: A good question 10, not hard, but lengthy.

Spoiler

So we know that:
x_A = -6 + 2t - t²/2
x_B = 4Sin[t]

Thus:

x'_A = v_A = 2 - t
x'_B = v_B = 4Cos[t]

Now do a quick graph of the velocites, and note that they cross twice, and only twice. [They can only cross twice because for t > 6, the velocity of A is increasing in the negative direction past an absolute value of 4]

To show they cross at least 2 times, let f(t) = v_B(t) - v_A(t) = 4Cos[t] + t - 2

f(0) = 4 + 0 - 2 = 2 > 0
f(pi) = 4Cos[pi] + pi - 2 = -6 + pi < 0
f(2pi) = 4 + 2pi - 2 = 2 + 2pi > 0

Thus there are at least 2 sign changes, and hence 2 places where f(t) = 0, and hence v_A = v_B.

To show there are no more than 2, we note that v_A cannot cross more than 1 peak of the cosine wave [as explained above], and that it begins below the original peak. Thus it passes through only 1 peak, and can have at most 2 crossings.

Thus; 2 <= crossings <= 2
I.e. crossings = 2

Wow, that was convoluted. But fairly rigoruos (for 2 unit level).
Technically showing it graphically should be enough.

To measure the distance a particle has travelled, it is not enough to measure it's change in displacement. So we need to find the turning points of the displacement function.

v_A = 0 = 2 - t
Thus t = 2 for the turning point (it's a parabola, no more working necessary, it's not a stationary point of inflexion as curvature cannot change)

Thus the total distance travelled is:
D = |x_A(t₁) - x_A(2)| + |x_A(t₂) - x_A(2)|
= x_A(2) - x_A(t₁) + x_A(2) - x_A(t₂)
= 2x_A(2) - (x_A(t₁) + x_A(t₂))

This is due to the sign of the velocity between the 2 areas, and thus the value of the displacement at t = 2 will always be greater than at any other place.

Note that 2x_A(2) = -12 + 4*2 - 2² = -8

Hence:

D = -8 - (-6 - 6 +2(t₁ + t₂) - (t₁² + t₂²)/2)
= 4 - 2(t₁ + t₂) + (t₁² + t₂²)/2)

As required.

Furthermore we have shown the maximum displacement of A is -4 @ t = 2, but alas, particle B @ t = 0 has:
x_B(2) = 3.63719

Thus the particles never meet [As the minimum displacement of particle B is -4, but not at that point.]

I'm awesome.

Next Question OF DOOM!

Rigorously prove Pythagorous' Theorem using at most 5 equals signs [Not including the equals signs within the congruence/gemoetry stage]. No skipping algebraic steps! [And you've got to rigorously prove the gemoetry, through construction etc.]

Alternatively, construct a set of cuts for an arbitrary square that resides on the hypotenuse of a right angled triangle, and reform 2 squares that can be placed on the opposite 2 sides.

 

[Jono_2007]

I said that someone else can do it i haven't covered the HSC topics yet, nor have i even started year 11 yet, but i like to have a go at some things, this looks to hard.
but: for the derivatives of the two particles you get the velocity.

Spoiler

So for particle a;
dx/dt=2-t and
for particle B;
dx/dt=4 cos(t)

Not sure if thats right but thats all i'll attempt!
Thanks for you help though.

 

[Jono_2007]

Oh sorry i didn't see your post.

 

[insert-username]

Rigorously prove Pythagorous' Theorem using at most 5 equals signs [Not including the equals signs within the congruence/gemoetry stage]. No skipping algebraic steps! [And you've got to rigorously prove the gemoetry, through construction etc.]

Spoiler

[Note: ‹ means "angle"]

1. Draw ▲ABC, with ‹A = 90°.

2. Construct AD perpendicular to BC (see attached diagram)

In ▲ABD and ▲ABC:

‹BAC = ‹BDA (right angles)
‹ABD = ‹ABC (common angle)
Therefore ‹BAD = ‹ACB (third angle)
Therefore ▲ABD ||| ▲ABC
Therefore AB/BC = BD/AB

i.e. AB² = BC.BD (1)

In ▲ACD and ▲ABC:

‹ADC = ‹BAC (right angles)
‹ACB = ‹ACD (common angle)
Therefore ‹DAC = ‹ABC (third angle)
Therefore ▲ACD ||| ▲ABC
Therefore AC/BC = DC/AC

i.e. AC² = BC.DC (2)

Add (1) and (2)

Therefore AB² + AC² = BC.BD + BC.DC

Therefore AB² + AC² = BC(BD + DC)

but (BD + DC = BC)

Therefore AB² + AC² = BC²

i.e. the square of the hypotenuse of a right-angled triangle is equal to the sum of the square of the other two sides.

Exactly 5 equals signs.

Next question:

Evaluate _-2∫³(2x + 1)² dx using:

i) Direct integration (2 unit techniques only)

ii) Simpson's rule with 3 function values


I_F

 

[Mountain.Dew]

Next question:

Evaluate _-2∫³(2x + 1)² dx using:

i) Direct integration (2 unit only)

ii) Simpson's rule with 5 function values


I_F

more of my 2 cents.

Spoiler

(i) _-2∫³(2x + 1)² dx = 1/6[(2x+1)³], from x=3 to x= -2

so, _-2∫³(2x + 1)² dx = 1/6( [2(3)+1]³ - 2(-2)+1]³)= 1/6(343 + 27) = 370/6

(ii) using Simpson's Rule -->

_-2∫³(2x + 1)² dx = 1/6[3-(-2)][f(3) + f(-2) + 4[f(1/2)]]
= 5/6[[2(3)+1]² + [2(-2)+1]² + 4[2(1/2) + 1]²]
=5/6[49 + 9 + 16] = 370/6

 

[insert-username]

more of my 2 cents.

Spoiler

(i) _-2∫³(2x + 1)² dx = 1/6[(2x+1)³], from x=3 to x= -2

so, _-2∫³(2x + 1)² dx = 1/6( [2(3)+1]³ - 2(-2)+1]³)= 1/6(343 + 27) = 370/3

(ii) using Simpson's Rule -->

_-2∫³(2x + 1)² dx = 1/6[3-(-2)][f(3) + f(-2) + 4[f(1/2)]]
= 5/6[[2(3)+1]² + [2(-2)+1]² + 4[2(1/2) + 1]²]
=5/6[49 + 9 + 16] = 370/6

Spoiler

I think you should check to see why you got 370/3 for one answer and 370/6 for the other. But for the record, 370/6 is correct for both answers.

Feel free to post up the next question.


I_F

 

[Mountain.Dew]

heres one from the 2002 Mathematics HSC

Question 7(a)

Consider the geometric series

1 + sqrt(5) - 2 + (sqrt(5) - 2)² + ...

(i) Explain why the geometric series has a limiting sum.
(ii) Find the exact value of the limiting sum. Write your answer with a rational denominator.
my additions:
(iii) evaluate the series for the first 20 terms.

 

[insert-username]

heres one from the 2002 Mathematics HSC

Question 7(a)

Consider the geometric series

1 + sqrt(5) - 2 + (sqrt(5) - 2)² + ...

(i) Explain why the geometric series has a limiting sum.
(ii) Find the exact value of the limiting sum. Write your answer with a rational denominator.
my additions:
(iii) evaluate the series for the first 20 terms.

Spoiler

(i) (√5 - 2) is less than 1, but greater than 0, so the series has a limiting sum.

(ii) S_∞ = a/(1-r)

= 1/[1 - (√5 - 2)]

= 1/(3-√5)

= (3+√5)/4


(iii) Do I have to?!

1, √5 -1, 6-3√5, can't be bothered doing any more....

I_F

 

[Riviet]

Is it just me or are the integral signs appearing as random letters?